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Complicated declarations in c

Many times you might have wondered about all the complex and strange syntaxes of functions, variables and pointers in c. But these declarations are not as complex as it looks at first. When you know where to look and how to look, reading these syntaxes would become a piece of cake. Just see this presentation and you will find it very easy to read all the complexities in c.

3.
Declaration of Pointers.
General Syntax:
[name of data type] *<variable_name>;
Here [name of data type] is data type of variable, address
of which this pointer can contain.
But it is not as simple as, above syntax is seems like.
All the keywords placed before ‘*’ does not affect
properties (they define properties of variable which can be
pointed by this pointer) of pointer variable. If we need to
affect properties of pointer variable all keywords must be
placed after * sign(but there are exceptions like far,huge
and near).

4.
Declaration of Pointers.
There are pointer type in c, like int * is a type of int
pointer.
If we assume int * is a type, and we know that we can
create pointers of any type by writing ‘[type name] * ‘.
So if we write int* * then it will create pointer of a int
pointer.
We can define a pointer to any other pointer in c.

7.
Const Phenomenon
What is the difference between following two declarations ?
const int *p;
int const *p;
Answer:
There is no difference, both refers to same meaning.

8.
Const Phenomenon
What is the difference between following two declarations ?
const int *p;
int * const p;
Answer:
First is a pointer to a constant int(*p=10 is wrong). And
second is a constant pointer to int.(p=10 is wrong).

10.
near,far and huge pointers
near,far and huge are used with pointers.
•These keywords were introduced to support memory
segmentation.
•Size of a near pointer is 2 bytes, and size of huge
and far pointers is 4 bytes.
•Near far and huge pointers only supported by
Borland tc under dos, 32 bit platforms like window,
linux and unix does not support these keywords,
there are all pointers are 32 bit.

11.
near,far and huge pointers
Which is the following is a near pointer ?
char near * far *ptr;
char near * huge *ptr2;
char far * huge* ptr3;
Answer:
No one is near pointer;

13.
Using () operator in declarations.
Like any other instruction of c, part of declarative
instruction contained by () operator executes separately,
and () operator defined by inner most ‘()’(level) executes
first and then one who has higher level of nesting.
Syntex:
Exp1(Exp2….(ExpN <variable_name>));
Exp1 ,Exp…ExpN are all sub expressions of a declarative
statement.

14.
Using [] operator in declarations
[] operator in a declaration define size of memory by
multiplying size of data type specified in statement with the
size specified in [] operator.
But what will happen with this memory size it depends on
declaration.

15.
Using [] operator in declarations
Observe following two declarations.
int *p[10];
int (*p)[10];
First one allocates 20 bytes and second one allocates two bytes.
First is a array of 10 int pointers.
And second is a pointer to a block which has size of 20 bytes.
In first statement [10] will going to be read first by
compiler. So compiler thought p as a array of 10 elements
and then it looks up for type of these elements which it
finds int *.
In Second statement () will executes first by compiler, so
compiler thought p as a pointer and then it looks up for
type of this pointer which it finds int [10].

17.
Use of () And [] in function pointer declarations.
We can have pointers to functions in c.
Read following declaration.
int *p(int,int);
Above is a declaration. Of a function named p with two int
parameters and int * as return type.
How can we declare a pointer to this function ?
Answer:
int * (*p)(int,int);
Now p is a pointer to a function which has two int parameters
and int *as return type.

18.
Use of () and [] in function pointer declarations.
Now, how we can declare a array of pointers of following function
int *p(int,int);
int (*(p[10]))(int,int);
In above declaration p is a array of 10 pointers of a function which
has two int parameters and int * as return type.
Now you can see, above is not the general declaration syntax of
pointers, as we saw earlier
[name of data type] *<variable_name>;
Now finally we came to some real complicated declarations.

20.
Rules to read declaration in c
1. Start with the inner most () containing variable
name.
2. Read the declarations clock wise. Read left first
in the same () and then right.
3. Read from inner most () to outer most().

22.
Rules to read declarations in c.
Char ( * ( * f () ) [] ) ();
F is function which returns a pointer to an array of pointers to
a function which returns an char.

23.
Can you read following declaration ?
void (*f())(void(*)(int *,void**),int(*)
(void**,int*));Answer:
Yes
In above declaration f is a function which returns a pointer to a
function which returns nothing and receives two parameters. First
is a pointer to a function which returns nothing but receives two
parameters first is pointer to an int and second is a pointer to a void
pointer. The second parameter is a pointer to a function which
retruns an int an receives two parameters first is pointer to a void
pointer and second is a pointer to a int.

24.
Can you declare a array of pointers of a
function which receives four
parameters all of following type and
returns following pointer.
void (*f())(void(*)(int *,void**),int(*)
(void**,int*));
Answer:
Do I Look like a fool ?

26.
How typedef works.
What are the types of var1,var2,var3 and var4 in following program ?
#include<stdio.h>
#define character char *
Void main()
{
Typedef char * CHARAT
character var1,var2;
CHARAT var3,var4;
}
Answer:
In above code var1,var3,var4 are pointers to char and
var2 is char variable.

27.
Will following code compile?
typedef struct
{
int data;
NODEPTR next;
} *NODEPTR;
Answer: No
A typedef declaration can not be used until it is defined,
and in following example it not defined when it is used.

29.
Can you declare a array of pointers of a function
which receives four parameters all of following
type and returns following pointer.
void (*f())(void(*)(int *,void**),int(*)
(void**,int*));
Answer:
typedef void (*f())(void(*)(int *,void**),int(*)(void**,int*));
f(*g[10])(f,f,f,f);

34.
Read following declaration.
char (*(*x[3])())[5];
Answer:
x is an array of 3 function pointers, where each pointer points to a
function that returns a pointer to array of 5 chars.

35.
Read following declaration.
void (*f[10])(int,int);
Answer:
f is an array of 10 function pointers, where each pointer points to a
function that receives two ints and returns nothing.

36.
Read following declaration.
int (*ftable[])(void) ={fadd,fsub,fmul,fdiv};
Answer:
Ftable is an array of 5 function pointers which points to the function
fadd(),fsub(),fmul(),fdiv()

37.
Read following declaration.
int ** (*f)(int **,int **(*)(int **,int **));
Answer:
f is a pointer to a function which returns a pointer to int pointer and
receives two parameters . First is a pointer of pointer of int. And
second is a pointer to a function which receives two pointers of int
pointer as parameters and returns a pointer to int pointer.

38.
What do the following declaration means ?
Typedef char *pc;
Typedef pc fpc();
Typedef fpc *pfpc;
Typedef pfpc fpfpc();
Typedef fpfpc *pfpfpc;
Pfpfpc a[10];
pc is a pointer to char.
fpc is a function which returns a pointer to char.
pfpc is a pointer to a function returning pointer to a char.
fpfpc is a function returning pointer to a function returning
pointer to a char.
pfpfpc is a pointer to function returning pointer to a function
returning pointer to char.
pfpfpc a[10] is an array of 10 pointers to a function returning
pointers to functions returning pointer to characters.